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Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition

Statement

Let $T : V \to W$ be a linear map between finite-dimensional real or complex inner product spaces, and let $r$ be the number of positive singular values of $T$ . Then there exist orthonormal bases $(e_{1}, \dots, e_{n})$ of $V$ and $(f_{1}, \dots, f_{m})$ of $W$ , together with singular values

$s_{1} \geq \dots \geq s_{r} > 0,$

such that

$T v = \sum_{j = 1}^{r} s_{j} ⟨ v, e_{j} ⟩ f_{j}$

for every $v \in V$ .

Facts & Assumptions

Given: A linear map $T : V \to W$ between finite-dimensional real or complex inner product spaces.

[L1]

The singular values of $T$ are the eigenvalues of the non-negative operator $∣ T ∣ = \sqrt{T^{*} T}$ (The singular values of a linear map as the eigenvalues of the positive square root of T^*T, Singular values are well defined because the positive square root of T^*T is unique).

[L2]

A non-negative operator has an orthonormal eigenbasis with nonnegative eigenvalues (A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S).

[L3]

Every finite-dimensional real or complex inner product space has an orthonormal basis (Every finite-dimensional real or complex inner product space has an orthonormal basis).

[L4]

A subspace of a finite-dimensional inner-product space and its orthogonal complement form an internal direct sum (For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ ).

Proof

technique · direct

1.1L1L2algebra

By [L1] and [L2], the non-negative operator $∣ T ∣$ on $V$ has an orthonormal eigenbasis $(e_{1}, \dots, e_{n})$ with $∣ T ∣ e_{j} = s_{j} e_{j}$ and $s_{1} \geq \dots \geq s_{r} > 0 = s_{r + 1} = \dots = s_{n}$ . Because $∣ T ∣^{2} = T^{*} T$ , one has $⟨ T^{*} T e_{j}, e_{j} ⟩ = s_{j}^{2}$ and therefore $∥ T e_{j} ∥ = s_{j}$ .

2.1step 1.1algebra

For each $j \leq r$ , define $f_{j} : = s_{j}^{- 1} T e_{j}$ . Then $∥ f_{j} ∥ = 1$ , and for $i \neq j$ one has $⟨ f_{i}, f_{j} ⟩ = (s_{i} s_{j})^{- 1} ⟨ T^{*} T e_{i}, e_{j} ⟩ = (s_{i} s_{j})^{- 1} s_{i}^{2} ⟨ e_{i}, e_{j} ⟩ = 0$ . For $j > r$ , step 1.1 gives $∥ T e_{j} ∥ = s_{j} = 0$ , so $T e_{j} = 0$ .

3.1L3L4step 2.1choosealgebra∎

Put $F = span (f_{1}, \dots, f_{r})$ , allowing $F = 0$ when $r = 0$ . By [L4], $W = F \oplus F^{⊥}$ . Choose an orthonormal basis of $F^{⊥}$ by [L3] and append it to $(f_{1}, \dots, f_{r})$ . The combined list is orthonormal because the two subspaces are orthogonal, and it spans $W$ by the direct-sum decomposition. It is therefore an orthonormal basis $(f_{1}, \dots, f_{m})$ of $W$ . If $W = 0$ , both lists are empty; if $r = 0$ , the complement basis alone is the required basis. If $v = \sum_{j = 1}^{n} ⟨ v, e_{j} ⟩ e_{j}$ , then step 2.1 gives $T v = \sum_{j = 1}^{n} ⟨ v, e_{j} ⟩ T e_{j} = \sum_{j = 1}^{r} s_{j} ⟨ v, e_{j} ⟩ f_{j}$ , and when $r = 0$ this is the empty sum $0$ .

Depends on

Dependency tree · 2 levels

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition